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Chapter 9 - 1

ISSUES TO ADDRESS... When we combine two elements...

what equilibrium state do we get? In particular, if we specify...

--a composition (e.g., wt% Cu - wt% Ni), and--a temperature ( T )

then...How many phases do we get?What is the composition of each phase?How much of each phase do we get?

Chapter 9: Phase Diagrams

Phase BPhase A

Nickel atomCopper atom

Chapter 9 - 2

Phase Equilibria: Solubility LimitIntroduction

Solutions solid solutions, single phase Mixtures more than one phase

Solubility Limit :Max concentration forwhich only a single phasesolution occurs.

Question: What is thesolubility limit at 20 C?

Answer: 65 wt% sugar .If C o < 65 wt% sugar: syrupIf C o > 65 wt% sugar: syrup + sugar.

65

Sucrose/Water Phase Diagram

P u r e

S u g a r

T e m p e r a t u r e

( C )

0 20 40 60 80 100C o =Composition (wt% sugar)

L(liquid solution

i.e., syrup)

SolubilityLimit L

(liquid)+S

(solidsugar)20

40

60

80100

P u r e

W a t e r

Adapted from Fig. 9.1,Callister 7e.

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Chapter 9 - 3

Components :The elements or compounds which are present in the mixture

(e.g., Al and Cu) Phases :

The physically and chemically distinct material regionsthat result (e.g., and ).

Aluminum-CopperAlloy

Components and Phases

(darkerphase)

(lighterphase)

Adapted fromchapter-openingphotograph,Chapter 9,Callister 3e.

Chapter 9 - 4

Effect of T & Composition ( C o ) Changing T can change # of phases:

Adapted fromFig. 9.1,Callister 7e.

D (100 C,90)2 phases

B (100 C,70)1 phase

path A to B . Changing C o can change # of phases: path B to D .

A (20 C,70)2 phases

70 80 1006040200

T e m p e r a

t u r e

( C )

C o =Composition (wt% sugar)

L(liquid solution

i.e., syrup)

20

100

40

60

80

0

L(liquid)+S

(solidsugar)

water-sugarsystem

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Chapter 9 - 5

Phase Equilibria

0.12781.8FCCCu

0.12461.9FCCNi

r (nm)electronegCrystalStructure

Both have the same crystal structure (FCC) and havesimilar electronegativities and atomic radii ( W. Hume Rothery rules ) suggesting high mutual solubility.

Simple solution system (e.g., Ni-Cu solution)

Ni and Cu are totally miscible in all proportions.

Chapter 9 - 6

Phase Diagrams Indicate phases as function of T , C o , and P . For this course:

-binary systems: just 2 components.-independent variables: T and C o (P = 1 atm is almost always used).

PhaseDiagramfor Cu-Nisystem

Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys , P. Nash(Ed.), ASM International, Materials Park,OH (1991).

2 phases:L (liquid) (FCC solid solution)

3 phase fields:LL +

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T (C)

L (liquid)

(FCC solidsolution)

L + l i q u

i d u s

s o l i d u

s

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Chapter 9 - 7

Note:The Ni-Cu alloy system shown in the previous slide is a binary isomorphous system;

Note: Isomorphous : Having the same structure. In the phase diagram sense,isomorphicity means having the same crystal structure or complete solid solubility for all compositions.

Chapter 9 - 8

wt% Ni20 40 60 80 10001000

1100

1200

1300

14001500

1600T (C)

L (liquid)

(FCC solidsolution)

L +

l i q u i d u s

s o l i d u

sCu-Niphase

diagram

Phase Diagrams :# and types of phases

Rule 1: If we know T and C o , then we know:--the # and types of phases present.

Examples:A(1100 C, 60):

1 phase: B (1250 C, 35):

2 phases: L +

Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys , P. Nash(Ed.), ASM International, Materials Park,OH, 1991).

B ( 1 2 5 0 C

, 3 5 )

A (1100

C,60)

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Chapter 9 - 9

wt% Ni20

1200

1300

T (C)

L (liquid)

(solid) L

+

l i q u i d u s

s o l i d u s

30 40 50

L +

Cu-Nisystem

Phase Diagrams :composition of phases

Rule 2: If we know T and C o , then we know:--the composition of each phase.

Examples:T A A

35C o32C L

At T A = 1320 C:Only Liquid ( L)C L = Co ( = 35 wt% Ni)

At T B = 1250 C:Both and LC L = C liquidus ( = 32 wt% Ni here)C = C solidus ( = 43 wt% Ni here)

At T D = 1190 C:Only Solid ( )C = C o ( = 35 wt% Ni )

C o = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys , P. Nash (Ed.), ASM

International, Materials Park, OH, 1991.)

B T B

D T D

tie line

4C 3

Chapter 9 - 10

Rule 3: If we know T and C o , then we know:--the amount of each phase (given in wt%).

Examples:

At T A: Only Liquid (L)W L = 100 wt%, W = 0

At T D : Only Solid ( )W L = 0, W = 100 wt%

C o = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys , P. Nash (Ed.), ASMInternational, Materials Park, OH, 1991.)

Phase Diagrams :weight fractions of phases

wt% Ni20

1200

1300

T (C)

L (liquid)

(solid) L

+

l i q u i d u s

s o l i d u s

30 40 50

L +

Cu-Nisystem

T A A

35C o

32C L

B T B

D T D

tie line

4C 3

R S

At T B : Both and L

%7332433543 wt =

=

= 27 wt%

W L =S

R + S

W =R

R + S

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Chapter 9 - 11

Tie line connects the phases in equilibrium witheach other - essentially an isotherm

The Lever Rule

How much of each phase?Think of it as a lever (teeter-totter)

M L M

R S

R M S M L=

L

L

LL

LL C C

C C S R

R W

C C C C

S R S

M M M

W

=+

=

=+

=+

=

00

wt% Ni

20

1200

1300

T (C)

L (liquid)

(solid) L

+

l i q u i d u s

s o l i d u s

30 40 50

L +

B T B

tie line

C o C L C

S R

Adapted from Fig. 9.3(b),Callister 7e.

Chapter 9 - 12

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T (C)

A

35C o

L: 35wt%Ni

Cu-Nisystem

Phase diagram:Cu-Ni system.

System is:--binary

i.e. , 2 components:

Cu and Ni.--isomorphousi.e., completesolubility of onecomponent inanother; phasefield extends from0 to 100 wt% Ni.

Adapted from Fig. 9.4,Callister 7e.

ConsiderC o = 35 wt%Ni .

Ex: Cooling in a Cu-Ni Binary

4635

4332

: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Ni

: 36 wt% Ni

B : 46 wt% NiL: 35 wt% Ni

C

D

E

24 36

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Chapter 9 - 13

C changes as we solidify. Cu-Ni case:

Fast rate of cooling:Cored structure

Slow rate of cooling:Equilibrium structure

First to solidify has C = 46 wt% Ni.Last to solidify has C = 35 wt% Ni.

Cored vs Equilibrium Phases

First to solidify:46 wt% Ni

Uniform C :35 wt% Ni

Last to solidify:< 35 wt% Ni

Chapter 9 - 14

Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on:

--Tensile strength ( TS ) --Ductility (% EL,% AR )

--Peak as a function of C o --Min. as a function of C o

Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.

T e n s

i l e S t r e n g t h

( M P a )

Composition, wt% NiCu Ni0 20 40 60 80 100

200

300

400

TS forpure Ni

TS for pure Cu E l o n g a

t i o n

( % E L )

Composition, wt% NiCu Ni0 20 40 60 80 10020

30

40

50

60

%EL for

pure Ni

%EL for pure Cu

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Chapter 9 - 15

: Min. melting T E

2 componentshas a special compositionwith a min. melting T.

Adapted from Fig. 9.7,Callister 7e.

Binary-Eutectic Systems

Eutectic transitionL(C E ) (C E ) + (C E )

3 single phase regions(L, , )

Limited solubility: : mostly Cu: mostly Ag

T E : No liquid below T E C E

composition

Ex.: Cu-Ag systemCu-Agsystem

L (liquid)

L + L+

+

C o , wt% Ag20 40 60 80 1000

200

1200T (C)

400

600

800

1000

C E

T E 8.0 71.9 91.2779 C

Chapter 9 - 16

L+ L+

+

200

T (C)

18.3

C , wt% Sn20 60 80 1000

300

100

L (liquid)

183 C61.9 97.8

For a 40 wt% Sn-60 wt% Pb alloy at 150 C, find...--the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (1)

+ --compositions of phases:

C O = 40 wt% Sn

--the relative amountof each phase:

150

40C o

11C

99C

S R

C = 11 wt% SnC = 99 wt% Sn

W =C - C O C - C

= 99 - 4099 - 11 =5988 = 67 wt%

S R +S =

W =C O - C C - C

=R R +S

= 2988

= 33 wt%= 40 - 1199 - 11

Adapted from Fig. 9.8,Callister 7e.

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Chapter 9 - 17

L+

+

200

T (C)

C , wt% Sn20 60 80 1000

300

100

L (liquid)

L+

183 C

For a 40 wt% Sn-60 wt% Pb alloy at 200 C, find...--the phases present: Pb-Sn

system

Adapted from Fig. 9.8,Callister 7e.

EX: Pb-Sn Eutectic System (2)

+ L--compositions of phases:

C O = 40 wt% Sn

--the relative amountof each phase:

W =C L - C O C L - C

=46 - 4046 - 17

=6

29 = 21 wt%

W L =C O - C C L - C

=2329 = 79 wt%

40 C o

46C L

17C

220S R

C = 17 wt% SnC L = 46 wt% Sn

Chapter 9 - 18

C o < 2 wt% Sn Result:

--at extreme ends

--polycrystal of grainsi.e., only one solid phase.

Adapted from Fig. 9.11,Callister 7e.

Microstructuresin Eutectic Systems: I

0

L+ 200

T (C)

C o , wt% Sn10

2

20C o

300

100

L

30

+

400

(room T solubility limit)

T E (Pb-SnSystem)

L

L: C o wt% Sn

: C o wt% Sn

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Chapter 9 - 19

2 wt% Sn < C o < 18.3 wt% Sn Result:

Initially liquid + then alonefinally two phases

polycrystalfine -phase inclusions

Adapted from Fig. 9.12,Callister 7e.

Microstructures

in Eutectic Systems: II

Pb-Snsystem

L +

200

T (C)

C o , wt% Sn10

18.3

200C o

300

100

L

30

+

400

(sol. limit at T E )

T E

2(sol. limit at T room )

L

L: C o wt% Sn

: C o wt% Sn

Chapter 9 - 20

C o = C E Result: Eutectic microstructure (lamellar structure)

--alternating layers (lamellae) of and crystals.

Adapted from Fig. 9.13,Callister 7e.

Microstructuresin Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.160 m

Micrograph of Pb-Sneutectic

microstructurePb-Snsystem

L +

+

200

T (C)

C , wt% Sn

20 60 80 1000

300

100

L

L+183 C

40

T E

18.3

: 18.3 wt%Sn

97.8

: 97.8 wt% Sn

C E 61.9

L: C o wt% Sn

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Chapter 9 - 21

Lamellar Eutectic Structure

Adapted from Figs. 9.14 & 9.15, Callister 7e.

Chapter 9 - 22

18.3 wt% Sn < C o < 61.9 wt% Sn Result: crystals and a eutectic microstructure

Microstructuresin Eutectic Systems: IV

18.3 61.9

S R

97.8

S R

primary eutectic

eutectic

W L = (1- W ) = 50 wt%

C = 18.3 wt% Sn

C L = 61.9 wt% SnS

R + S W = = 50 wt%

Just above T E :

Just below T E :C = 18.3 wt% SnC = 97.8 wt% Sn

S R + S

W = = 73 wt%

W = 27 wt%Adapted from Fig. 9.16,Callister 7e.

Pb-Snsystem

L+200

T (C)

C o , wt% Sn

20 60 80 1000

300

100

L

L+

40

+

T E

L: C o wt% Sn LL

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Chapter 9 - 23

L+ L+

+

200

C o , wt% Sn20 60 80 1000

300

100

L

T E

40

(Pb-SnSystem)

Hypo eutectic & Hyper eutectic

Adapted from Fig. 9.8,Callister 7e. (Fig. 9.8adapted from Binary Phase Diagrams , 2nd ed., Vol. 3,T.B. Massalski (Editor-in-Chief), ASM International,Materials Park, OH, 1990.)

160 meutectic micro-constituent

Adapted from Fig. 9.14,Callister 7e.

hypereutectic: (illustration only)

Adapted from Fig. 9.17,Callister 7e. (Illustrationonly)

(Figs. 9.14 and 9.17from Metals Handbook , 9th ed.,Vol. 9,Metallography and

Microstructures ,American Society forMetals, MaterialsPark, OH, 1985.)

175 m

hypoeutectic: C o = 50 wt% Sn

Adapted fromFig. 9.17, Callister 7e.

T (C)

61.9eutectic

eutectic: C o =61.9wt% Sn

Chapter 9 - 24

Intermetallic Compounds

Mg2Pb

Note: intermetallic compound forms a line - not an area -because stoichiometry (i.e. composition) is exact.

Adapted fromFig. 9.20, Callister 7e.

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Chapter 9 - 25

Eutectoid & Peritectic Eutectic( ) - liquid in equilibrium with two solids

L + coolheat

intermetallic compound- cementite

coolheat

Eutectoid( ) - solid phase in equation with twosolid phasesS 2 S 1+S 3

+ Fe 3C (727C)

coolheat

Peritectic( ) - liquid + solid 1 solid 2 (Fig 9.21)

S 1 + L S 2 + L (1493C)

Chapter 9 - 26

Eutectoid & Peritectic

Cu-Zn Phase diagram

Adapted fromFig. 9.21, Callister 7e.

Eutectoid transition +

Peritectic transition + L

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Chapter 9 - 27

Iron-Carbon (Fe-C) Phase Diagram 2 important

points

-Eutectoid ( B ): + Fe 3C

-Eutectic ( A):L + Fe 3C

Adapted from Fig. 9.24, Callister 7e .

F e 3

C ( c e m e n

t i t e )

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe 3C

+Fe 3C

+

L+Fe 3C

(Fe) C o , wt% C

1148 C

T (C)

727 C = T eutectoid

AS R

4.30Result: Pearlite =alternating layers of and Fe 3C phases

120 m

(Adapted from Fig. 9.27, Callister 7e .)

R S

0.76

C e u

t e c t o i

d

B

Fe 3C (cementite-hard) (ferrite-soft)

Chapter 9 - 28

Hypo eutectoid Steel

Adapted from Figs. 9.24and 9.29, Callister 7e .(Fig. 9.24 adapted fromBinary Alloy Phase Diagrams , 2nd ed., Vol.1, T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)

F e 3

C ( c e m e n t

i t e )

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe 3C

+ Fe 3C

L+Fe 3C

(Fe) C o , wt% C

1148 C

T (C)

727 C

(Fe-CSystem)

C 0 0

. 7 6

Adapted from Fig. 9.30, Callister 7e .proeutectoid ferritepearlite

100 m Hypoeutectoidsteel

R S

w =S /(R +S )w Fe 3C =(1- w )

w pearlite = w pearlite

r s

w =s /(r +s )w =(1- w )

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Chapter 9 - 29

Hyper eutectoid Steel

F e 3

C ( c e m e n t

i t e )

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe 3C

+Fe 3C

L+Fe 3C

(Fe) C o , wt%C

1148 C

T (C)

Adapted from Figs. 9.24and 9.32, Callister 7e .(Fig. 9.24 adapted fromBinary Alloy Phase Diagrams , 2nd ed., Vol.1, T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)

(Fe-CSystem)

0 . 7 6 C o

Adapted from Fig. 9.33, Callister 7e .

proeutectoid Fe 3C

60 mHypereutectoidsteel

pearlite

R S

w =S /(R +S )w Fe 3C =(1- w )

w pearlite = w pearlite

s r

w Fe 3C =r /( r +s )w =(1- w Fe 3C )

Fe 3C

Chapter 9 - 30

Example: Phase Equilibria

For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine thefollowing

a) composition of Fe 3C and ferrite ( )b) the amount of carbide (cementite) in grams

that forms per 100 g of steelc) the amount of pearlite and proeutectoid

ferrite ( )

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Chapter 9 - 31

Chapter 9 Phase EquilibriaSolution:

g3.94g5.7CFe

g7.5100022.07.6022.04.0

100xCFe

CFe

3

CFe3

3

3

=

=

=

=

=+

x

C C C C o

b) the amount of carbide(cementite) in grams thatforms per 100 g of steel

a) composition of Fe 3C and ferrite ( )C O = 0.40 wt% CC = 0.022 wt% CC Fe C = 6.70 wt% C3

F e 3

C ( c e m e n

t i t e )

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe 3C

+ Fe 3C

L+Fe 3C

C o , wt% C

1148 CT (C)

727C

C O

R S

C Fe C3C

Chapter 9 - 32

Chapter 9 Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ( )

note: amount of pearlite = amount of just above T E

C o = 0.40 wt% CC = 0.022 wt% CC

pearlite= C

= 0.76 wt% C

+

=C o C C C

x 100 = 51.2 g

pearlite = 51.2 gproeutectoid = 48.8 g

F e 3

C ( c e m e n

t i t e )

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe 3C

+ Fe 3C

L+Fe 3C

C o , wt% C

1148 CT (C)

727 C

C O

R S

C

C

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Chapter 9 - 33

Alloying Steel with More Elements

T eutectoid changes: C eutectoid changes:

Adapted from Fig. 9.34, Callister 7e . (Fig. 9.34from Edgar C. Bain, Functions of the Alloying Elements in Steel , American Society for Metals,1939, p. 127.)

Adapted from Fig. 9.35, Callister 7e . (Fig. 9.35from Edgar C. Bain, Functions of the Alloying Elements in Steel , American Society for Metals,1939, p. 127.)

T E u t e c

t o i d ( C )

wt. % of alloying elements

Ti

Ni

Mo Si W

Cr

Mn

wt. % of alloying elements

C e u

t e c t o i

d ( w t % C )

Ni

Ti

Cr

SiMn

WMo

Chapter 9 - 34

Phase diagrams are useful tools to determine:--the number and types of phases,--the wt% of each phase,--and the composition of each phase

for a given T and composition of the system. Alloying to produce a solid solution usually

--increases the tensile strength ( TS )--decreases the ductility.

Binary eutectics and binary eutectoids allow fora range of microstructures.

Summary

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Chapter 9 - 35

Core Problems:

Self-help Problems:

ANNOUNCEMENTSReading: